Answer to Question #288458 in Statistics and Probability for M Umair Farooq

Question

Draw all possible distinct sample of size 3 from the population 2 ,4, 6, 8, 10, 12,

Construct the sampling distribution of the sample mean X and calculate it's means and variance meauX and sigma X

Answer 1

Mean

$$\mu=\dfrac{2+4+6+8+10+12}{6}=7$$

Variance

$$\sigma^2=\dfrac{1}{6}\big((2-7)^2+(4-7)^2+(6-7)^2$$ $$+(8-7)^2+(10-7)^2+(12-7)^2\big)$$ $$=\dfrac{70}{6}=\dfrac{35}{3}$$

Standard deviation

$$\sigma=\sqrt{\sigma^2}=\sqrt{\dfrac{35}{3}}\approx 3.415650$$

We have population values $2,4,6,8,10,12$ population size $N=6$ and sample size $n=3.$ Thus, the number of possible samples which can be drawn without replacement is

$$\dbinom{6}{3}=\dfrac{6!}{3!(6-3)!}=20$$ $$\def\arraystretch{1.5} \begin{array}{c:c:c} Sample & Sample & Sample \ mean \\ No. & values & (\bar{X}) \\ \hline 1 & 2,4,6 & 4 \\ \hdashline 2 & 2,4,8 & 14/3 \\ \hdashline 3 & 2,4,10 & 16/3 \\ \hdashline 4 & 2,4,12 & 6 \\ \hdashline 5 & 2,6,8 & 16/3 \\ \hdashline 6 & 2,6,10 & 6 \\ \hdashline 7 & 2,6,12 & 20/3 \\ \hdashline 8 & 2,8,10 & 20/3 \\ \hdashline 9 & 2,8,12 & 22/3 \\ \hdashline 10 & 2,10,12 & 8 \\ \hdashline 11 & 4,6,8 & 6 \\ \hdashline 12 & 4,6,10 & 20/3 \\ \hdashline 13 & 4,6,12 & 22/3 \\ \hdashline 14 & 4,8,10 & 22/3 \\ \hdashline 15 & 4,8,12 & 8 \\ \hdashline 16 & 4,10,12 & 26/3 \\ \hdashline 17 & 6,8,10 & 8 \\ \hdashline 18 & 6,8,12 & 26/3 \\ \hdashline 19 & 6,10,12 & 28/3 \\ \hdashline 20 & 8,10,12 & 10 \\ \hdashline \end{array}$$

The sampling distribution of the sample means.

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f & f(\bar{X}) & \bar{X}f(\bar{X})& \bar{X}^2f(\bar{X}) \\ \hline 4 & 1& 1/20 & 12/60 & 144/180 \\ \hdashline 14/3 & 1 & 1/20 & 14/60 & 196/180 \\ \hdashline 16/3 & 2 & 2/20 & 32/60 & 512/180 \\ \hdashline 6 & 3 & 3/20 & 54/60 & 972/180\\ \hdashline 20/3 & 3 & 3/20 & 60/60 & 1200/180 \\ \hdashline 22/3 & 3 & 3/20 & 66/60 & 1452/180 \\ \hdashline 8 & 3 & 3/20 & 72/60 & 1728/180 \\ \hdashline 26/3 & 2 & 2/20 & 52/60 & 1352/180 \\ \hdashline 28/3 & 1 & 1/20 & 28/60 & 784/180 \\ \hdashline 10 & 1 & 1/20 & 30/60 & 900/180 \\ \hdashline Total & 20 & 1 & 420/60 & 9240/180 \\ \hline \end{array}$$

$$E(\bar{X})=\sum\bar{X}f(\bar{X})=\dfrac{420}{60}=7$$

The mean of the sampling distribution of the sample means is equal to the

the mean of the population.

$$E(\bar{X})=\mu_{\bar{x}}=7=\mu$$

$$Var(\bar{X})=\sum\bar{X}^2f(\bar{X})-(\sum\bar{X}f(\bar{X}))^2$$ $$=\dfrac{9240}{180}-(7)^2=\dfrac{7}{3}$$ $$\sigma_{\bar{X}}=\sqrt{Var(\bar{X})}=\sqrt{\dfrac{7}{3}}\approx1.527525$$

Verification:

$$Var(\bar{X})=\dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})=\dfrac{35}{3(3)}(\dfrac{6-3}{6-1})$$ $$=\dfrac{7}{3}, True$$